(a) Define uniform acceleration. (b) Forces act on a car in motion. List the (i) horizontal forces and their directions; (ii) vertical forces and their dire...

(a) Uniform acceleration. Uniform acceleration is a constant rate of change of velocity with time, that is, the velocity of the body changes by equal amounts in equal intervals of time.

(b) Forces acting on a moving car.

(i) Horizontal forces and their directions:

• The driving (tractive) force from the engine through the wheels, acting forward in the direction of motion.
• Friction between the tyres and road together with air resistance (drag), acting backward, opposite to the direction of motion.

(ii) Vertical forces and their directions:

• The weight of the car, acting vertically downward.
• The normal reaction from the road surface, acting vertically upward.

(c)(i) Velocity-time graph. The car speeds up uniformly from rest to \(25\text{ m s}^{-1}\) in the first \(20\text{ s}\) (line \(OA\)), holds \(25\text{ m s}^{-1}\) for the next \(30\text{ s}\), that is until \(t=50\text{ s}\) (line \(AB\)), then slows uniformly to rest by \(t=60\text{ s}\) (line \(BC\)).

(c)(ii)(\(\alpha\)) Total distance travelled. The distance is the area under the graph, which is the trapezium \(OABC\). The parallel sides are the whole base \(OC=60\text{ s}\) and the top \(AB=30\text{ s}\); the height is the speed \(25\text{ m s}^{-1}\):

\[ s=\tfrac{1}{2}(OC+AB)\times v=\tfrac{1}{2}(60+30)\times 25=\tfrac{1}{2}\times 90\times 25=1125\text{ m.} \]

(c)(ii)(\(\beta\)) Deceleration. The deceleration is the magnitude of the slope of the final line \(BC\), where the speed falls from \(25\text{ m s}^{-1}\) to \(0\) in the last \(10\text{ s}\) (from \(t=50\text{ s}\) to \(t=60\text{ s}\)):

\[ a=\frac{\text{change in velocity}}{\text{time}}=\frac{0-25}{60-50}=\frac{-25}{10}=-2.5\text{ m s}^{-2}. \]

So the car decelerates at \(2.5\text{ m s}^{-2}\). Note the value is \(2.5\text{ m s}^{-2}\), not \(25\text{ m s}^{-2}\): the change of \(25\text{ m s}^{-1}\) is divided by the \(10\text{ s}\) taken.

Work done on the spring (force-extension graph). For a spring obeying Hooke's law the work done, which is the elastic potential energy stored, equals the area under the force-extension line. From the graph the line runs from the origin to \(F=12\text{ N}\) at an extension \(e=0.5\text{ cm}=0.5\times10^{-2}\text{ m}=0.005\text{ m}\), so the area is a triangle:

\[ W=\tfrac{1}{2}\,F\,e=\tfrac{1}{2}\times 12\times 0.005=0.03\text{ J.} \]

The work done on the spring is \(0.03\text{ J}\), that is \(3.0\times10^{-2}\text{ J}\). Remember to convert the extension from centimetres to metres before multiplying, otherwise the energy comes out one hundred times too large.